Is Height of Special Binary Tree easy or hard?

The problem is rated Medium because the traversal itself is straightforward, but recognizing and ignoring the circular leaf connections can be tricky. Once those special pointers are filtered, the remaining logic becomes a standard binary tree height computation.

Height of Special Binary Tree Python/Java solution

Most implementations use a DFS helper function. The function recursively computes left and right subtree heights while ignoring circular leaf links. The same logic works across Python, Java, C++, Go, and TypeScript with time complexity O(n) and recursion space O(h).

How to solve Height of Special Binary Tree in O(n)?

Run a DFS starting from the root and compute subtree heights recursively. Before traversing a child, check whether the pointer is part of the circular leaf connection (node.left.right == node or node.right.left == node). Skip those edges so the traversal follows only real parent-child relationships. This guarantees O(n) time since each node is processed once.

What is the best approach for Height of Special Binary Tree?

Depth-First Search (DFS) is the most common approach. Traverse the tree recursively while ignoring the circular links between leaf nodes by checking if node.left.right == node or node.right.left == node. This visits each node once and computes the height using 1 + max(left, right). The total complexity is O(n) time and O(h) space.

Is Height of Special Binary Tree asked at Google/Amazon/Meta?

Tree traversal and height calculation problems frequently appear in interviews at companies like Amazon, Google, and Meta. Variants that include structural constraints—such as circular leaf links or modified pointers—are common follow-up questions to test deeper understanding of tree traversal logic.

What data structure is used in Height of Special Binary Tree?

The problem uses a binary tree where each node has left and right pointers. The special twist is that leaf nodes are connected in a circular doubly linked list using those same pointers. Solutions typically use recursion with DFS or a queue for BFS to traverse the tree structure safely.

What is the time complexity of Height of Special Binary Tree?

The optimal solution runs in O(n) time because every node in the tree is visited exactly once during traversal. DFS uses O(h) space for the recursion stack where h is the tree height, while a BFS approach uses O(w) space where w is the maximum width of the tree.

Description Hints 4 Solution Video 8:05

2773. Height of Special Binary Tree

Medium75.2% acceptancePremium elsewhere, free here

You are given a root, which is the root of a special binary tree with n nodes. The nodes of the special binary tree are numbered from 1 to n. Suppose the tree has k leaves in the following order: b₁<b₂< ... < b_k.

The leaves of this tree have a special property! That is, for every leaf b_i, the following conditions hold:

The right child of b_i is b_{i + 1} if i < k, and b₁ otherwise.
The left child of b_i is b_{i - 1} if i > 1, and b_k otherwise.

Return the height of the given tree.

Note: The height of a binary tree is the length of the longest path from the root to any other node.

Example 1:

Input: root = [1,2,3,null,null,4,5]
Output: 2
Explanation: The given tree is shown in the following picture. Each leaf's left child is the leaf to its left (shown with the blue edges). Each leaf's right child is the leaf to its right (shown with the red edges). We can see that the graph has a height of 2.

Example 2:

Input: root = [1,2]
Output: 1
Explanation: The given tree is shown in the following picture. There is only one leaf, so it doesn't have any left or right child. We can see that the graph has a height of 1.

Example 3:

Input: root = [1,2,3,null,null,4,null,5,6]
Output: 3
Explanation: The given tree is shown in the following picture. Each leaf's left child is the leaf to its left (shown with the blue edges). Each leaf's right child is the leaf to its right (shown with the red edges). We can see that the graph has a height of 3.

Constraints:

n == number of nodes in the tree
2 <= n <= 10⁴
1 <= node.val <= n
The input is generated such that each node.val is unique.

Topics

Hints open one at a time

Hint 1

To solve the problem, we must first distinguish leaves from internal nodes.

Hint 2

For some node v, if v.left == null or v.right == null, then v is not a leaf.

Hint 3

If the previous condition does not hold, and v.left.right == v, then v is a leaf node.

Hint 4

Now that we can check if some node is a leaf, we can make the function “heightOfTree” a recursive function that returns the tree's height in which its input is the root of that subtree.

8:05

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1class Solution:
  def solve(self):
      pass


Starter codePython 3Spaces: 4

Examples

Case 1Case 2Case 3

Input

[1,2,3,null,null,4,5]

Input

[1,2]

Input

[1,2,3,null,null,4,null,5,6]

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